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The problem asks to find \( \frac{dz}{dt} \), where:

\[
z = xy + x^2y
\]
and
\[
x = at^2, \quad y = 2at.
\]

We need to differentiate \( z \) with respect to \( t \), using the given expressions for \( x \) and \( y \).

### Step 1: Expand \( z \)
First, let's rewrite \( z \) more clearly:
\[
z = xy + x^2y.
\]
Now substitute \( x = at^2 \) and \( y = 2at \) into this equation.

\[
z = (at^2)(2at) + (at^2)^2(2at)
\]
\[
z = 2a^2t^3 + 2a^3t^5.
\]

### Step 2: Differentiate \( z \) with respect to \( t \)
Now, differentiate \( z = 2a^2t^3 + 2a^3t^5 \) with respect to \( t \).

\[
\frac{dz}{dt} = \frac{d}{dt}(2a^2t^3) + \frac{d}{dt}(2a^3t^5)
\]
\[
\frac{dz}{dt} = 6a^2t^2 + 10a^3t^4.
\]

Thus, the derivative of \( z \) with respect to \( t \) is:

\[
\boxed{\frac{dz}{dt} = 6a^2t^2 + 10a^3t^4.}
\]

Would you like a detailed explanation of any step, or do you have any specific questions?

Here are some related questions you could explore:
1. What happens if \( y = 3at \) instead of \( y = 2at \)?
2. How would the result change if \( x = at^3 \) instead of \( x = at^2 \)?
3. Can you find the second derivative, \( \frac{d^2z}{dt^2} \), from this expression?
4. What is the physical interpretation of \( z = xy + x^2y \) in terms of motion?
5. How does the derivative change if the function \( z \) includes trigonometric terms for \( x \) or \( y \)?

**Tip:** Whenever you differentiate composite functions, remember to apply the chain rule carefully, especially when variables depend on \( t \).

Find dz/dt, when z = xy + x^2y and x = at^2, y = 2at.

This problem involves finding the derivative dz/dt where z = xy + x^2y, with x = at^2 and y = 2at. Learn the step-by-step process to apply the chain rule and product rule in this calculus example.

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